Dawn dead in Ceres orbit, ran out of fuel Oct 2018

[OmCheeto]

...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html
Emily says Dawn is approaching Ceres "from the south". That means we should be able to see more small-circle rotation detail at the bottom of the Ceres image. I think I do. This agrees with the upper Om-figure of post #62. Going by that figure, Dawn crosses the equatorial plane of Ceres around 15 Feb.
Or let's say the orbit plane determined by direction to sun (left) and orbit direction (into the page).
But the axial tilt is only about 3 deg, so the planes are nearly parallel. Until mid-February Dawn is looking slightly from the south and sees more of the south polar region. Increasingly after mid-Feb it will be looking more from the north.
Please correct if I'm missing something. In the animation, lighting also seems stronger near the north pole, with northern regions blanked out in glare. We are told the planet's axial tilt is only 3 degrees, could the current tilt (small as it is) be towards the sun?

Dr. Rayman, on Friday stated:

Marc Rayman says:
February 6, 2015 at 10:49 am
Tom,

You’re right that the direction of Ceres’ axis is not precisely known (yet!), but it is tipped only a few degrees from the plane of its orbit. (This might help others understand a bit more about the axial tilt.) It seems to me then that in one practical (albeit unconventional) sense, Ceres is always near equinox. That is, the sun is always near the equator, so, yes indeed, most of the surface is lit during a Cerean day. That would also mean the poles are never more than “just barely illuminated.” Astronomers have several estimates of the pole location but using one popular one, a colleague calculated for me that the sun right now is about 4 degrees south latitude and near the southern summer solstice.

I should add that Dawn is not over Ceres’ equator, so we are not seeing the northern and southern hemispheres equally. It is approaching over the southern hemisphere, and it was above (roughly) 22 degrees south latitude for OpNav 3. Dawn will cross the equator between RC1 and RC2.

Marc

My guess is that the washing out of the polar regions may be due to the lack of available light.

Current angular size: 2 arcsin(.475 / 110.67) = 0.491832996 deg (as of 9AM pacific, on 8 Feb)

Om has some more figures tabulated in post #55 but here is the table of optical navigation photo shoots (with a few extra rows added) from 29 January D.J.
http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/

Jan 25    (237,000)    43    (22)     1.3    96%
Feb 1     (168,000)
Feb 3     (146,000)    70    (14)     2.2    97%
Feb 10    (100,000)
Feb 12    (83,000)    121    (7.8)    3.8    98%
Feb 19    (46,000)    221    (4.3)    7.0    87%
Feb 25    (40,000)    253    (3.7)    8.0    44%
Mar 1     (49,000)    207    (4.6)    6.5    22%
Apr 10    (33,000)    304    (3.1)    9.6    18%
Apr 15    (22,000)    455    (2.1)    14     50%
Apr 23    (14,000)

Relevant column headings as given in the original table:
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk

In an effort to figure out what "Capture by Ceres' gravity" means, I spent the last few days doing some "orbital mechanics" studying.
The only thing I've concluded, is that Newton was a FREAK!
Even with spreadsheets, the internet, and one of the worlds most powerful laptops, I have not a clue what I'm doing.

But in my feeble attempt to solve the capture problem, I discovered lots of weird and wonderful things, which is always a good thing.

About the only useful thing I can add, is my z-y plot of Dawn, as it approaches Ceres, as viewed from the Sun. (z becoming the x-axis in the 2D plot below)

Ceres is at the origin.
Position labels are date and distance in kkm from my digitization.
3-6 & 3-20 are extractions from one of Dr. Rayman's journals.

 

[marcus]

My guess is that the washing out of the polar regions may be due to the lack of available light.

Good guess! I'm glad you got a response from Dr. Rayman about the axial tilt. Just a few degrees (Wikipedium on Ceres say about 3 deg) and it is currently Southern summer--around the solstice in fact.

Om your extra interest and energetic research makes all the difference! Keeps me feeling optimistic and excited by what we are learning in this thread.

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition. That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

 

[marcus]

About capture, I measured on your annotated diagram and it looks to me like in either diagram the distance from Ceres at time of capture is 40% more than the distance on 1 March which is 49000 km. So around 6 March, capture, distance must be R = 68000 or 69000 something like that.
So we can calculate the upper limit on speed (2GM/68000 km)^.5
remembering that the mass of our planet is M = 943 billion billion kg.
Just paste this into google: (2G*943e18 kg/68000 km)^.5
When I do that, google comes back with 43 m/s
Actually it says "((2 * G * (943e18 kg)) / (68 000 km))^.5 = 43.0233559 m / s" :D

Current status says the current speed is about 100 m/s. So if there is going to be capture on 6 March then by this rough calculation the speed relative to Ceres has to come down to below 43 m/s.

I didn't measure very accurately just held a plastic ruler up to the diagrams on the screen. It was about 40% farther on 6 March than on 1 March.
Current angular size: 2 arcsin(.475 / 108.83) = 0.500148519 deg (as of 3PM pacific, on 8 Feb)

BINGO, half a degree. Full moon size!

 

[marcus]

It's always fun to see better and better pictures, but I would say that the next big landmark day for me will be FEBRUARY 24 WHEN DAWN IS 28 kkm directly North and

28 kkm directly behind Ceres in their race around the sun.

so that the hypoteneuse distance works out to 40 kkm (28²+28²=40²).

The idea is that 28 thousand km lag should be the FARTHEST BEHIND CERES THAT DAWN EVER TRAILS.

It has to cling fiercely to Ceres' coat-tails (if Grain&Fertility godesses can be said to have coat tails).
It has to stop drifting behind at that point and from that day forward begin catching up.

The combined x,y,z distance will generally be greater than that but from then on the behind distance along the Ceres orbit track has to be gradually reduced until they are neck and neck and
Dawn is safely in a circular polar orbit around the planet.

 

[OmCheeto]

Good guess! I'm glad you got a response from Dr. Rayman about the axial tilt. Just a few degrees (Wikipedium on Ceres say about 3 deg) and it is currently Southern summer--around the solstice in fact.

Om your extra interest and energetic research makes all the difference! Keeps me feeling optimistic and excited by what we are learning in this thread.

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition. That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

Well, I figured the speed had to be somewhere between the orbital and escape velocities. But one of the only relevant equations I found was for a Kepler orbit, which required a "semi-major axis", which, given my complete geometrical incompetence, is completely useless. How on Earth do you extract a "semi-major axis" from a curve segment?

I think I'll just sit back now, and wait for the pretty pictures.

 

[mfb]

Every speed below the escape velocity will give an orbit. We don't know the planned 3D motion so it is hard to estimate the speed of Dawn. I guess the speed relative to Ceres will drop below the escape velocity at that point. It will make Dawn the first object to have been in orbit around two different celestial objects (not counting Earth).

 

[marcus]

Om, However it happened we got fairly close agreement for the speed V around "capture" on 6 March. I was just roughly estimating and got 43 m/s and you got 45 m/s I think.
Whatever the speed actually is, if capture occurs that day then the speed must be just slightly LESS than escape speed (43 or 45, something around there).

 

[lpetrich]

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition.

Yes, that's a reasonably good approximation, though it's a spherical-cow sort of formula. Or, in this case, a spherically-symmetric asteroid all alone.

I'll estimate the sizes of the two main sources of departure from that ideal state: Ceres's equatorial bulge and the Sun. Ceres should have an equatorial bulge from its rotation, and one can estimate its size from the hypotheses of hydrostatic equilibrium and constant density.

Ceres's equatorial bulge:
$$ \text{Oblateness gravitational coefficient } J_2 \sim \text{flattening } f \sim \left( \frac{T_{SS}}{T_{rotation}} \right)^2 $$
SS = surface satellite

For Ceres, it should be about 0.06 or 1/17. To do a better job, one should use the formulas for Maclaurin spheroids, as they are called.
$$ \text{EB's relative effect } \sim J_2 \left( \frac{r_{equatorial}}{r} \right)^2 $$

The Sun:
$$ \text{Sun's relative effect} \sim \frac{M_{Sun}}{M} \left( \frac{r}{r_{to\ Sun}} \right)^3 \sim \left( \frac{T_{SS}}{T_{Sun\ orbit}} \right)^2 \left( \frac{r}{r_{equatorial}} \right)^3 $$

At 100,000 km (100 megameters):
EB: 1.4*10^(-6)
Sun: 0.026

That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

A good approximation ought to be constant acceleration, since Dawn's ion engines have *very* low thrust, and since they are typically run almost continuously.

 

[mfb]

The uncertainty on the mass of Ceres is much larger than the effect of the equatorial bulge.

Dawn has a constant acceleration as measured by Dawn, but not in the system of Ceres.
103400km now (90m/s relative velocity), which means Ceres appears larger to Dawn than the moon to us (using the semimajor axis).

Oh, and Ceres is a very good approximation to a spherical cow in frictionless vacuum. It does not give milk, however.

 

[OmCheeto]

Every speed below the escape velocity will give an orbit. We don't know the planned 3D motion

I kind of do.

The x-axis would be a line from the sun to Ceres.
The y-axis would be through the north pole.
The z-axis is along the line of motion of Ceres.

I derived this from plotting out the points from the published images.
Much massaging has been done, and I do not recommend piloting crafts in this manner.
The original y and z values jumped around so badly, that it looked like pinball.
Then I found a couple of buttons on my spreadsheet: Trendlines & Polynomial

so it is hard to estimate the speed of Dawn.

It will be fun to see how close I got, just by eyeballing the graphs.

I guess the speed relative to Ceres will drop below the escape velocity at that point.

I'm afraid I still don't understand how this works.
But as wiki states; "The consequences of the rules of orbital mechanics are sometimes counter-intuitive."
I have found this to be quite true.
When I first saw stated that PE tends towards zero at infinite distances, I scoffed.
Everyone here one Earth knows the equation is PE = mgh, and therefore, as h approaches infinity, so should PE.
Things are different, out in space.

It will make Dawn the first object to have been in orbit around two different celestial objects (not counting Earth).

Yay!

 

[marcus]

Since we just turned a page I will bring forward the essentials. Here's Emily Lakdawalla's version of the latest shots.
...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla

Here's Om's date/distance labeled version of Marc Rayman's approach trajectory diagrams. In upper, the trajectory path is projected onto the plane normal to Ceres orbit motion. So Ceres' motion is into the page. The upper diagram does not show how the probe falls behind Ceres (almost 30 thousand km) until 25 Feb, when it begins to catch up. Because that falling behind would be out of the page. The lower diagram (which Dawn entered on the left edge today, 10 Feb) shows the path projected down on Ceres orbit PLANE, with the sun off to the left and the planet's orbit motion directly up on the page. This is the same path projected on a different plane. It shows how the probe falls behind at first by nearly 30 thousand km--I estimate 28,000 km but that remains to be seen.

Here's the link to current status:
http://neo.jpl.nasa.gov/orbits/fullview2.jpg

 

[marcus]

25 Feb is an important upcoming date because it is when the spacecraft will hopefully stop falling behind Ceres and (with Ceres' gravity helping) begin to catch up. At that time, I estimate that Dawn will be 28 kkm behind Ceres (measured on the orbit plane projection) and 28 kkm up measured on the upper diagram---the projection on the plane orthogonal to planet motion. Combining those two distances at right angles to each other gives 40 kkm, which agrees with Marc Rayman's figure for 25 Feb.

http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/

Jan 25    (237,000)    43    (22)     1.3    96%
Feb 1     (168,000)
Feb 3     (146,000)    70    (14)     2.2    97%
Feb 10    (100,000)
Feb 12    (83,000)    121    (7.8)    3.8    98%
Feb 19    (46,000)    221    (4.3)    7.0    87%
Feb 25    (40,000)    253    (3.7)    8.0    44%
Mar 1     (49,000)    207    (4.6)    6.5    22%
Apr 10    (33,000)    304    (3.1)    9.6    18%
Apr 15    (22,000)    455    (2.1)    14     50%
Apr 23    (14,000)

Relevant column headings as given in the original table, which listed planned navigational photo shoots. (I added three rows, without photo data, to help connect with the trajectory diagrams.)
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk

 

[marcus]

Current status is for 11 Feb at 4AM UTC, which is 10 Feb at 8 PM pacific time. (8 hours earlier than UTC)
The distance given is 93.02 kkm (thousand kilometers).
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
2 arcsin(.475 / 93.02) = 0.585... deg
117% the size of a full moon.

Those estimates I gave for 25 Feb are just rough approximations. I said 28 kkm up (off the orbit plane) and trailing 28 kkm behind (for a combined distance of 40 kkm.
But it could, for instance be a bit less up and a bit more back.
It could be say 27.1 kkm up and 29.4 kkm back

The main thing is that around 25 Feb it stops falling behind and begins to catch up to Ceres in their race around the sun, and the farthest it falls behind is something short of 30 thousand kilometers.

If you have been periodically checking the current status page, with the simulated view of Ceres seen from Dawn, you may be enjoying seeing that brown dot in the middle of the frame grow large.
The white dot you get as a cursor, on that page, is full moon size (it's a 30 degree field of view and the dot is 1/60 of the frame width). So it gives something to compare with.

 

[marcus]

Current status ( http://neo.jpl.nasa.gov/orbits/fullview2.jpg ) now shows the Dawn spacecraft in picture-taking mode--having temporarily stopped thrusting.
And having rotated so as to change orientation--point the camera I guess.
So we can expect a bunch of new photographs (perhaps in a day or two, after processing) unless they are just calibrating instruments this time.

This particular photo-shoot is designated RC1. There should be a noticeable improvement in resolution over last time (assuming we get to see the results.)

Current status says distance as of 7PM pacific on 11 Feb (3h 12Feb UTC) is 86.43 kkm
2 arcsin(.475 / 86.43) = 0.6298 deg
126% of full moon
It gives the approach speed as 90 m/s. That's around 8 thousand km a day. So it is reasonable to suppose that some of the pictures, maybe all, will be shot at the nominal range of 83 kkm .
==========
Looking ahead, on 19 Feb, and 25 Feb, range will be 46 kkm and 40 kkm so angular size will be
2arcsin(.475/46) and 2arcsin(.475/40) respectively
237% and 272% of full moon size, respectively.

 

[marcus]

Om, I recall you checked out the Deep Space Network (DSN) to see if it was receiving images from Dawn, at one point, during a past photo shoot.
I got curious and tried to do this. I saw three locations---Goldstone, Madrid, Canberra. Each location had several antennas.
http://eyes.nasa.gov/dsn/dsn.html
I see! you click on the number under each antenna and it tells you which spacecraft the antenna is receiving data from. At the moment none of the antennas were taking Dawn data.

EDIT: I checked more recently (9PM pacific time) and saw that Dawn was talking to Canberra. Also an antenna at Madrid appeared to be standing ready to take over.

Checked current status as of 5PM pacific 12 Feb, distance 80.03 thousand km
2 arcsin(.475 / 80.03) = 0.680136325 deg
136% of moon-size :D

 

[OmCheeto]

Om, I recall you checked out the Deep Space Network (DSN) to see if it was receiving images from Dawn, at one point, during a past photo shoot.
I got curious and tried to do this. I saw three locations---Goldstone, Madrid, Canberra. Each location had several antennas.
http://eyes.nasa.gov/dsn/dsn.html
I see! you click on the number under each antenna and it tells you which spacecraft the antenna is receiving data from. At the moment none of the antennas were taking Dawn data.

EDIT: I checked more recently (9PM pacific time) and saw that Dawn was talking to Canberra. Also an antenna at Madrid appeared to be standing ready to take over.

Checked current status as of 5PM pacific 12 Feb, distance 80.03 thousand km
2 arcsin(.475 / 80.03) = 0.680136325 deg
136% of moon-size :D

I checked at around 5:30 pm PST yesterday, and communication had started.
This morning, around 2 am PST, communication had by that time, switched to Madrid.
Currently, there is no communication, and I see that Dawn's ion drive is back on. (Duh!)

I should mention, that my mathematical skills have become quite rusty over the last 30 years, from dis-use. But the plotting, and mathematical magic, of Dawn's trajectory have really re-sparked my interest. Not to mention, that I've discovered 3D rendering software on my laptop.

I discovered that it can even create an animation of the above. But at 20 megabytes, it's a bit above my storage limit.

 

[mfb]

Two images from different viewing angles, switching back and forth, would be sufficient to get a nice 3D impression.

 

[OmCheeto]

Two images from different viewing angles, switching back and forth, would be sufficient to get a nice 3D impression.

Never mind. I just checked my account. I'm not even close to my limit. :)

But my latest attempt came out to 34 megabytes! It took me 8 minutes to upload.

3D Animation

 

[lpetrich]

I've measured the positions off of marcus's most recent diagram. They are as a tab-delimited spreadsheet text file that I've attached to this message.

Columns:
Date
Distance in kkm, where given
X1 raw -- first picture horizontal
Y1 raw -- first picture vertical
X2 raw -- second picture horizontal
Z2 raw -- second picture vertical

The second row is my measurement of Ceres's position in the picture

Scaling my measured positions to Dawn's distances from Ceres has proved more tricky than I expected, so I'll use Mathematica for that.

BTW, OmCheeto, what software did you use for:
- Measuring Dawn's position off of marcus's picture
- Plotting Dawn's position in 3D
?

I wrote an image measurer for myself since I couldn't find a good one that enters a position with each click on the picture being measured. It's OSX-native, so to port it to Windows or Linux, you'll need GNUstep.

 

[marcus]

Om,
Impressive bit of animation, thanks for sharing it. The figure in your post #87 rotates! I'm curious: what type of account puts a storage limit on this? Is it a PF limit, or one connected with your ISP (internet service provider)? Or has it to do with some "cloud" thing that you upload to, analogous to YouTube for videos or SoundCloud for music.

 

[marcus]

LPetrich,
I printed off your numbers, and I am trying to judge the difference in scale. For Ceres the upper X1 and lower X2 are 238 and 321
that is, in the lower image the planet is farther to the right. So if I take Ceres as my origin, or zero, then the farthest right the probe goes is around 20 March
Say 582 - 238 = 344 on the upper
and 560 - 321 = 239 on the lower
As I interpret it, this determines the relative scale of the two diagrams. that is "239" on the lower is the same distance as "344" on the upper diagram.
Does this make sense to you?

You are doing the work, so maybe I should not make suggestions. But my inclination would be to subtract off the Ceres coordinates and make Ceres the origin. And then multiply the lower diagram distances by 344/239 to scale them up to be the same size as in the upper diagram. I wonder if this seems reasonable to you?

BTW the originals of the two diagrams are in the November Dawn Journal, as far as I know.
http://dawnblog.jpl.nasa.gov/2014/11/28/dawn-journal-november-28/
The date labels are figured out from taking the capture date to be 6 March, which is what Marc Rayman estimated it would be. this is marked on both trajectories

 

[lpetrich]

Those are raw pixel coordinates, right off the images. I did it that way so that it's easy to compare with others' measurements. Yes, it's a raw-data release. :)

My measurement of Ceres is in the second row. So you can subtract that out, at least if you think that it's not grossly in error.

I found that the first diagram has a true distance per pixel about 3/2 the second diagram, without a large error. So my measurements of Ceres's position in the pictures is not far off. I also found that 1 kkm = 3.22 second-diagram pixels, also without great error. I guess I should upload my adjusted version.

 

[OmCheeto]

Om,
Impressive bit of animation, thanks for sharing it. The figure in your post #87 rotates! I'm curious: what type of account puts a storage limit on this? Is it a PF limit, or one connected with your ISP (internet service provider)? Or has it to do with some "cloud" thing that you upload to, analogous to YouTube for videos or SoundCloud for music.

The limit is from my ISP. I've had my own webspace since around 1996.
I broke my 50 megabyte limit a while back from all the images I was posting to PF.
I didn't realize until today, that my $5/month upgrade bumped me up to a 750 mb limit.
I guess I'm still kind of stuck in my 4 kilobyte ram and 300 baud mindset, from when I got my first PC.

That little "movie" just bumped my usage from 90 mb to 120 mb.

 

[marcus]

Om, Petrich, Mfb,
as of right now Ceres is 50% larger than the full moon. This is the according to the current status report as of 10:30 PM pacific on 13 Feb
which is the same as 6:30 AM UT on 14 Feb. this is as I am posting this
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
says distance is 72.73 kkm which corresponds to 0.75 degree or 150% of moon's angular size

I'm impatient to see the photos that were taken yesterday.

 

[OmCheeto]

Om, Petrich, Mfb,
as of right now Ceres is 50% larger than the full moon. This is the according to the current status report as of 10:30 PM pacific on 13 Feb
which is the same as 6:30 AM UT on 14 Feb. this is as I am posting this
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
says distance is 72.73 kkm which corresponds to 0.75 degree or 150% of moon's angular size

I'm impatient to see the photos that were taken yesterday.

My guess is, that they are going to wait until 1 minute after midnight, such that it will be a Valentines day present.
Set your alarm, and take a nap! :D
That's what I'm doing.

 

[lpetrich]

Standard gravitational parameter - Wikipedia

In celestial mechanics, the standard gravitational parameter μ of a celestial body is the product of the gravitational constant G and the mass M of the body.

μ = GM

For several objects in the Solar System, the value of μ is known to greater accuracy than either G or M. The SI units of the standard gravitational parameter are m³s⁻².

For Ceres, it is 63.1(3) km³s⁻², or an error of 1/20.

I've included my calculations in my latest attachment. It's another TSV text file, and I've zipped it to save space.

In addition to my raw positions, I've calculated:
Positions in pixels with Ceres's position subtracted out
Scaled and combined 3D positions - kkm
Distances for them -- kkm
Smoothed 3D positions (5-point with quadratic fit) -- kkm
Distances for them -- kkm
3D velocities (2-point differences) -- m/s
Velocity magnitudes -- m/s
Orbital velocities -- m/s
Escape velocities -- m/s

I get capture at March 5 - 6, and the last velocities are close to the circular-orbit velocities for their distances.

 

[marcus]

Thanks Petrich! I just printed it out and will have a look. Almost midnight here, so may not get back to you until tomorrow.
It looks like you ran a successful check because capture worked out right (around 6 March) and the final orbit velocities are right.
I don't understand what you did very well at present, but I think it will become clearer after a little time.

 

[mfb]

I used your numbers to calculate accelerations from gravity and from Dawn. And while the second derivative of those numbers is not very reliable, the data points are good enough for an interesting result: gravitational acceleration from Ceres exceeds the average acceleration from Dawn's ion drives during the fly-by end of this month, probably 2-3 days before Feb 20 where you have the first velocity data.

Expressed in (Mm)^2*m/(s*day), GM=5450. That leads to 3m/(s*day) at 42.5km distance, while I get ~2m/(s*day) average for the ion drives, increasing to 4m/(s*day) close to RC3 (but there I would need a better integration scheme to get more reliable values).

Quick cross-check: Using the values of 3000 I_sp and 1 kW estimated power I found, continuous thrust would allow ~5800kg*m/s momentum change per day. Dawn is lighter than 1000 kg now, so those velocity changes look reasonable.
Also, most of the time the calculated ion drive acceleration is against the direction of motion, something you would expect from an enery/fuel-saving approach.

 

[lpetrich]

I used for smoothing a quadratic function fitted to 5 points with equal weight. That resulted in this filter kernel: (1/35)*{-3,12,17,12,-3}

For derivatives, I used this filter kernel: (1/2)*{-1,0,1}

I calculated Dawn's orbital energy, and it declines until about March 17, then it levels off until about April 7, then it starts declining again. Since Dawn's ion engines are typically run for long stretches of time, I conclude that there will be a gap in its running between March 17 and April 7, a gap that will let Dawn fall toward Ceres.

Orbital energy:
$$ E = \frac12 v^2 - \frac{\mu}{r} $$
Strictly speaking, energy per unit spacecraft mass. It changes at this rate:
$$ \frac{dE}{dt} = {\mathbf v} \cdot {\mathbf a} $$
for acceleration a. Doing the case of going from one circular orbit to another in an orbit that is as circular as possible, I find:

(Final orbital velocity) - (initial orbital velocity) = +- (propulsion delta-V) (+ for inward, - for outward)

Also, Dawn will get to about 40 kkm at about February 23, giving Ceres an angular size of 1.5 degrees or 3 times the Moon's from the Earth. It will then move away to about 80 kkm at about March 18, with Ceres being 0.75d across or 1.5 times the Moon's.. Late in April, it should be less than 20 kkm away, giving an angular size greater than 3d or 6 times the Moon's.

Dawn should reach its lowest Ceres-relative velocity at about March 21, a little more than 10 m/s (36 km/h, 22 mph). It is currently traveling at 90 m/s (320 km/h, 200 mph) relative to Ceres. Late in April, it should get up to about 60 m/s (220 km/h, 130 mph).

It will spend some time in a high-altitude orbit before going into a low-altitude orbit. The lowest possible one is for a surface satellite, and that sort of satellite will have orbital velocity 360 m/s (1300 km/h, 800 mph) and period 2.3 hours. So after getting into a high orbit, Dawn will expend about 300 m/s delta-V for getting into a low orbit. But it's not expected to do any more traveling, since escaping Ceres will require a similar amount of delta-V and going to some other sizable object will require even more delta-V.

I've uploaded the Mathematica notebook that I've been working on for these calculations.

 

[OmCheeto]

Just for the record, I fully approve of your numbers, lpetrich.

Ceres and Dawn are moving "into" the image.

Dawn and Ceres are moving "up" in the image​

Except for the apex, our numbers are very close.
And I should mention that your apex numbers are better than mine.
I think I massaged mine a bit too much, as my raw data at that point matched yours much more closely.

I used for smoothing a quadratic function fitted to...

You apparently know your maths...
I, unfortunately, have lost too many brain cells to figure out how to interpret some of this "magic".

 

[marcus]

To get a rough idea of how it shapes up in 3D I have copied Petrich's X,Y,Z, and distance numbers (smoothed where available.) X is distance out from sun with Ceres X = 0, and Z is the lag behind Ceres, which at first increases until probe begins catching up. We want Z to be small because that means Dawn has caught up with Ceres Z = 0.
Y is a bit unintuitive, it is the distance BELOW the orbit plane (i.e. roughly in Ceres South pole direction) so when it becomes more negative that means the probe is rising in the North pole direction, above Ceres orbit plane Y = 0.

I will fill in some more of Petrich's numbers in this abbreviated table later as time permits.

date      X          Y          Z        distance from Ceres
F17   -45.9972    -6.4086    27.2882    53.86518747                  
F18   -38.555    -9.71627    28.2185    48.75630903
F19   -32.3324    -12.4392    29.202    45.30865911                  
F20   -26.169    -14.8491    29.9728    42.46976656
F21   -19.6171    -17.2648    30.4689    40.14047582
F22   -13.2794    -19.4975    30.6993    38.71617222
F23   -6.73346    -21.6416    30.593    38.07400666
F24   -0.502056    -23.4431    30.212    38.24390071
F25     5.62894    -25.0851    29.7158    39.29345973
F26     11.407    -26.4613    29.1488    40.98746867
F27     17.2899    -27.6663    28.1919    43.11783881
F28     22.8583    -28.5286    27.0313    45.46508635
M1      27.9985    -29.1842    25.6846    47.90962543
M2      32.8862    -29.7513    24.1873    50.51403254
M3      37.6439    -30.1647    22.7166    53.31994232
M4      41.9734    -30.4246    21.3167    56.05197848
M5      45.8274    -30.5605    19.8726    58.55779181
M6      49.5028    -30.6491    18.2955    61.02966376
M7      52.8252    -30.4896    16.7451    63.24963112
M8      55.7681    -30.3242    15.1946    65.27261258
M9      58.5427    -30.0761    13.6441    67.21592802
M10     58.5427    -30.0761    13.6441    67.21592802
M11     63.2886    -29.4796    10.5963    70.61710331
M12     65.1256    -29.1961    9.32939    71.97772953

 

[mfb]

As Dawn is within the Hill sphere of Ceres, the meaning of Z and Y do not matter much. Total distance matters. And X relative to the total distance, because it determines how much sun we see on the surface.

 

[lpetrich]

Except for the apex, our numbers are very close.
And I should mention that your apex numbers are better than mine.
I think I massaged mine a bit too much, as my raw data at that point matched yours much more closely.

Thanx. Good that two independent measurers agree.

(my smoothing algorithm)

You apparently know your maths...
I, unfortunately, have lost too many brain cells to figure out how to interpret some of this "magic".

It's a rather standard sort of procedure. For each point, find a smooth curve that goes through it and its neighbors. Then find the curve's value at the point's location.

If the curve is a linear function of its parameters, like a polynomial with its coefficients, and if one does least-squares fitting, then one can do much of the fitting work in advance. The smoothed point's value then becomes a weighted sum of its original value and its neighbors' values. One finds those weights by doing that advance fitting work.

 

[marcus]

Hi Mfb, glad you are contributing to the discussion!
Om and Petrich, thanks so much for the numerical and graphic work! It makes a big improvement.
Besides ourselves, I don't know who else might be reading. In case others are, I'm thinking that since we are studying one approach trajectory it might be interesting to make a comparison with the earlier one that was planned before the accident in September caused several days loss of thrust.

The second approach, developed after the accident, takes more than five weeks longer. So it shows what a difference a slight deficiency in orbit speed can make. Timing is sensitive.
In the first approach Dawn had all the necessary orbit speed so it was not lagging behind the planet, and the time-consuming gravity assist was not needed. This figure from the November Journal shows the comparison. In both cases "capture" occurs on 5-6 March. So you can count the day circles thereafter and see how much longer it takes to reach the initial target orbit (labeled RC3).

The sun is off to the left--Dawn approaches from sunward. Their common solar orbit direction is into the page.
In the first case, Dawn would have been up to speed and able to use thrust to slow down (short spacing between day circles as it enters diagram). So it slips right into RC3 orbit. Clean and direct
In the second case, Dawn is still needing its thrust to match Ceres' orbit speed, so it has not been able to slow down its sideways approach from sunward (long spacing between day circles as it enters) and moreover it is still falling behind as it approaches!
So it overshoots and uses Ceres' gravity to help slow down. It is also using Ceres' gravity to help it catch up. (the falling behind is up out of the page and not shown, nor is the crucial catching up, which is what takes such a long time).

These additional maneuvers, and the additional navigation photo shoots, consume the attitude control propellant (hydrazine) which has become the critical factor which could limit successful completion of the mission. Dawn was launched from Earth carrying a 45 kg supply of hydrazine and the mission team has had to budget that supply carefully. Planned navigational photoshoots have been canceled because they require rotating the orientation of the spacecraft .

Before the temporary loss of thrust in September, the planned arrival in RC3 orbit was 17 March.
After the necessary change of approach trajectory, projected arrival at RC3 will be over 5 weeks later.

BTW current status puts the range at 68.23 kkm at the moment (4PM pacific 14Feb, or midnight UT) making Ceres 160% of moonsize

 

[OmCheeto]

...
I wrote an image measurer for myself since I couldn't find a good one that enters a position with each click on the picture being measured. It's OSX-native, so to port it to Windows or Linux, you'll need GNUstep.

I was wondering how you did that so fast.
Is that X-code?
Not only do I need to relearn maths, I need to relearn how to code.